Decimal expansion of rational numbers

The factor (7^2) makes the decimal recurring, and the larger exponent among (2) and (5) is (3), giving the non-repeating start. In exams, separate recurrence from the initial delay.

Since (63=3^2\cdot 7), the reduced denominator is (2^4\cdot 5^3). The reason directly explains the terminating decimal rule.

(0.\overline{625}) is a fixed recurring decimal, so it is rational but not terminating. A decimal is terminating only when zeros continue after some point.

After cancellation, the fraction becomes (\frac{3}{2^2\cdot 5^6}). The larger exponent is (6), so the decimal terminates after (6) places.

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^6) means the larger exponent is (7) or (8).

(0.00\overline{72}=\frac{72}{9900}=\frac{2}{275}). The correct denominator is (275), so choose (275) from the options.

Two non-repeating zeros and two repeating digits give (\frac{72}{9900}). Reducing it gives (\frac{2}{275}).

After cancellation, the denominator becomes (2^4\cdot 5^3). The larger exponent is (4), so the decimal terminates after (4) places.

A decimal terminates when the reduced denominator has only (2) and (5). The other statements are incomplete because other prime factors may also be present.

When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus (0.124999\ldots=0.125).

The factors (3^2) and (17) must be removed from the reduced denominator, so the minimum factor is (3^2\cdot 17=153). Factors (2) and (5) may remain.

Since (189=3^3\cdot 7), the reduced denominator is (2^2\cdot 5). The larger exponent is (2), so the decimal terminates after (2) places.

This decimal does not end, and the number of zeros between the (2)'s keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Since (10^6=2^6\cdot 5^6), the denominator lacks (5^4). Thus (N=11\cdot 5^4=6875), so the correct option is (6875).

To obtain denominator \(10^6=2^6\cdot5^6\), we must make the power of 5 equal to 6. The given denominator is \(2^6\cdot5^2\), so multiply numerator and denominator by \(5^4=625\). Hence \(N=11\cdot5^4=11\cdot625=6875\). Common mistakes: 1375 equals \(11\cdot125\) (using \(5^3\) instead of \(5^4\)), while 2750 comes from another incorrect multiplier. Exam tip: prime-factorize the denominator and match exponents of 2 and 5 to form \(10^k\).

The reduced denominator is (10^5), so the decimal terminates exactly after (5) places. The numerator condition indicates no further cancellation.

In the first option, (121=11^2) cancels the denominator's (11), leaving only (2) and (5) in the denominator, so it terminates. No option is non-terminating here, so the options need rechecking.

In (\frac{49}{2\cdot 5^2\cdot 7^2}), (49=7^2) cancels completely, so it terminates. For a non-terminating recurring decimal, a factor other than (2) and (5) must remain in the reduced denominator.

A rational number has a terminating decimal expansion after the fraction is reduced only when the denominator has no prime factors other than 2 and 5. If any other prime factor remains in the lowest terms, its decimal expansion is non-terminating but recurring. Therefore, cancellation must be performed before classifying the decimal; looking only at the original denominator could give a wrong conclusion.

Here, \(14=2\cdot7\), so cancellation with the numerator gives \(\frac{14}{2\cdot5^2\cdot7^2}=\frac{1}{5^2\cdot7}\). The reduced denominator still contains the prime factor 7. Hence the decimal cannot terminate and must be non-terminating recurring. Thus option B is correct. It is not non-recurring because every rational number has either a terminating or recurring decimal expansion.

At most (5) decimal places means the fraction can be written with denominator (10^5). The reduced denominator must divide (10^5).